Let $R$ be the region enclosed by the polar curve $r(\theta)=2-2\sin(\theta)$, as shown in the graph. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{3\pi}{2}}\left( 2-4\sin(\theta)+2\sin^2(\theta)\right)d\theta$ (Choice B) B $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{3\pi}{2}}\left( 1-2\sin(\theta)+\sin^2(\theta)\right)d\theta$ (Choice C) C $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{5\pi}{2}}\left( 2-4\sin(\theta)+2\sin^2(\theta)\right)d\theta$ (Choice D) D $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{5\pi}{2}}\left( 1-2\sin(\theta)+\sin^2(\theta)\right)d\theta$
This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ We know $r(\theta)$ but we still need to figure out $\alpha$ and $\beta$. A nice starting point for the boundaries would be the point where the curve goes through $(0,0)$. The first non-negative $\theta$ -value for which $r(\theta)=0$ is $\dfrac{\pi}{2}$. So $\alpha=\dfrac{\pi}{2}$. $\beta$ is the next $\theta$ -value after $\alpha$ for which $r(\theta)=0$. That means $\beta=\dfrac{5\pi}{2}$. Let's plug ${r(\theta)=2-2\sin(\theta)}$, ${\alpha=\dfrac{\pi}{2}}$, and ${\beta=\dfrac{5\pi}{2}}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{\scriptsize\dfrac{\pi}{2}}}^{{\scriptsize\dfrac{5\pi}{2}}}\dfrac{1}{2}\left({2-2\sin(\theta)}\right)^{2}d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{5\pi}{2}}\dfrac{1}{2}\left( 4-8\sin(\theta)+4\sin^2(\theta)\right)d\theta \\\\ &= \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{5\pi}{2}}\left( 2-4\sin(\theta)+2\sin^2(\theta)\right)d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{5\pi}{2}}\left( 2-4\sin(\theta)+2\sin^2(\theta)\right)d\theta$